Question

If A(0,3,4), B(1,5,6), C(-2,0,-2) are the vertices of a triangle ABC and the bisector of angle A meets the side BC at D, then AD =

• A. \( \frac{\sqrt{21}}{5} \)
• B. \( \frac{\sqrt{42}}{10} \)
• C. 10
• D. 4

Hint:

Calculate coordinate differences mentally while applying the distance formula to save writing time.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

Using the Angle Bisector Theorem, point D divides BC in the ratio of the adjacent sides AB and AC. We calculate the lengths, find the coordinates of D, and then the length AD.
Step 2: Key Formula or Approach:

1. Distance formula: \( \sqrt{(x_2-x_1)^2 + \dots} \). 2. Section formula: \( D = \frac{mC + nB}{m+n} \) where \( m/n = AB/AC \).
Step 3: Detailed Explanation:

Calculate lengths of AB and AC: \( AB = \sqrt{(1-0)^2 + (5-3)^2 + (6-4)^2} = \sqrt{1 + 4 + 4} = 3 \). \( AC = \sqrt{(-2-0)^2 + (0-3)^2 + (-2-4)^2} = \sqrt{4 + 9 + 36} = 7 \). Ratio \( AB : AC = 3 : 7 \). D divides BC internally in ratio \( 3:7 \). Coordinates of D: \( x_D = \frac{3(-2) + 7(1)}{10} = \frac{1}{10} \) \( y_D = \frac{3(0) + 7(5)}{10} = \frac{35}{10} \) \( z_D = \frac{3(-2) + 7(6)}{10} = \frac{36}{10} \) Length AD: \( AD = \sqrt{(\frac{1}{10}-0)^2 + (\frac{35}{10}-3)^2 + (\frac{36}{10}-4)^2} \) \( AD = \sqrt{\frac{1}{100} + (\frac{5}{10})^2 + (\frac{-4}{10})^2} \) \( AD = \sqrt{\frac{1 + 25 + 16}{100}} = \sqrt{\frac{42}{100}} = \frac{\sqrt{42}}{10} \)
Step 4: Final Answer:

The length is \( \frac{\sqrt{42}}{10} \).